Due to surface tension, the excess pressure inside a smaller drop is $9$ units. If $27$ smaller drops combine then excess pressure inside the bigger drop is

Excess pressure inside a liquid drop of radius $r$ is given by $P = \dfrac{2T}{r}$.

For the smaller drop, $P_1 = \dfrac{2T}{r} = 9$ units.

When $27$ smaller drops combine to form a bigger drop of radius $R$, the total volume remains conserved.

$\dfrac{4}{3}\pi R^3 = 27 \times \dfrac{4}{3}\pi r^3$

$R^3 = 27r^3 \Rightarrow R = 3r$

The excess pressure inside the bigger drop is $P_2 = \dfrac{2T}{R}$.

Substituting $R = 3r$, we get:

$P_2 = \dfrac{2T}{3r} = \dfrac{1}{3} \left( \dfrac{2T}{r} \right)$

$P_2 = \dfrac{1}{3} \times 9 = 3$ units.

Answer: $3$ units